Is x+y-z=2 a subspace of R^3?

Question

Hello,

I am a little confused about this problem - I know that in order for something to be considered a subspace it must pass through the origin (and thus include the zero vector). Does this mean that x+y-z=2 is not a subspace of R^3?

Any help would be appreciated!

Answer 1

S = {(x, y, z) l x + y - z = 2}

0 + 0 - 0 ≠ 2 So, (0,0,0) is not in S. Therefore, S is not a subspace of R³.